# Gamma for ATM options with low spots

I'm trying to compute gamma for a vanilla call with spot and strike equal to 0.001. BLACK & SCHOLES formula gave me a value of 554.761 for gamma which is a very high. I have then two questions: Deos gamma correspond to the variation of Delta compared to 1% or 1 unit change of spot? (Actually, 1 unit spot shift is not relevant for small spots) What is the BS formula that gives gamma in case the variation is 1% of spot? Thanks in advance for your answer.

• Gamma, as the derivative of the delta with respect to underlying spot price, represents the rate of change in the delta under an infinitesimal variation in spot price. In analogy with physics, it’s the instantaneous acceleration of the option price. Nov 28, 2019 at 17:07
• gamma corresponds to the variation of delta in response to a one unit movement of the underlying stock price. The issue is for small values of spot, one unit shift is not relevant. It is more logical to use 1 percent shift so i'm looking for a modified formula in order to compute gamma value for 1percent spot shift. Nov 28, 2019 at 17:24

Gamma is the sensitivity of the delta with respect to infinitesimal changes in the price of the underlying asset (in whatever unit your underlying is nominated, typically dollar, pounds, euros, ...). So, it is not a percentage change. Instead, the percentage change (option elasticity) equals $$\Delta\frac{S}{V}$$. This quantity, for instance, gives you the expected excess return of the option.

However, it sounds to me as if you're seeking the elasticity of delta? The percentage change in delta when the price of the underlying asset changes by one percent? This is given by \begin{align*} \frac{\frac{\partial \Delta}{\Delta} }{\frac{\partial S}{S}} &= \frac{\partial \Delta }{\partial S}\frac{S}{\Delta} \\ &= \Gamma\frac{S}{\Delta}. \end{align*} In the Black-Scholes case, $$\Delta_c=e^{-qT}\Phi(d_1)$$ or $$\Delta_p=-e^{-qT}\Phi(-d_1)$$ and $$\Gamma=e^{-qT}\frac{\varphi(d_1)}{S\sigma\sqrt{T}}=Ke^{-rT}\frac{\varphi(d_2)}{S^2\sigma\sqrt{T}}$$, which is obviously the same for puts and calls. Here, $$\varphi$$ and $$\Phi$$ are the pdf and cdf of a standard normally distributed random variable.

If you were only interested in the absolute change in delta if the price of underlying asset changes by one percent, you would compute $$\frac{\partial \Delta}{\frac{\partial S}{S}}=\Gamma S$$ which is identical for European-style put and call options. Similarly, the percentage change in delta given an absolute change in the underlying asset (in the corresponding units) is given by $$\frac{\frac{\partial\Delta}{\Delta}}{\partial S}=\frac{\Gamma}{\Delta}$$.

## Edit

In response to Slade's important question, let me comment on the factor $$\frac{1}{100}$$. The equations above ignore units and just look at the ratio of absolute/relative changes in price/delta ... If you divide two percentages, you ''lose'' the percentage symbol and ought to divide by 100 again. So the factor $$\frac{1}{100}$$ is just to produce a number like between 0.012 whereas otherwise you get 1.2[%] where you need to remember the %. So the numbers are the same, it's up to mere preference.

Example: Black Scholes world with $$S_\mathrm{old}=10$$, $$K=10$$, $$r=0.05$$, $$T=\frac{1}{2}$$ and $$\sigma=0.2$$ (no dividends).

Then, $$C_\mathrm{old}\approx 0.69$$, $$\Delta_\mathrm{old}\approx0.60$$ and $$\Gamma_\mathrm{old}\approx0.27$$.

Consider now an absolute change in the price of the underlying asset to $$S_\mathrm{abs}=11$$ such that $$C_\mathrm{abs}\approx1.41$$, $$\Delta_\mathrm{abs}\approx 0.82$$ and $$\Gamma_\mathrm{abs}\approx0.17$$.

Similarly, we consider a percentage change to $$S_\mathrm{per}=10.1$$ with $$C_\mathrm{per}\approx0.75$$, $$\Delta_\mathrm{per}\approx 0.62$$ and $$\Gamma_\mathrm{per}\approx 0.27$$.

So, what do we get with these numbers?

• Trivially, $$S_\mathrm{abs}\approx C_\mathrm{old} + \Delta_\mathrm{old}=1.29$$ or even better $$S_\mathrm{abs}\approx C_\mathrm{old}+\Delta_\mathrm{old}+\frac{1}{2}\Gamma_\mathrm{old}=1.42$$. Similarly, $$\Delta_\mathrm{abs}\approx \Delta_\mathrm{old}+\Gamma_\mathrm{old}=0.87$$.
• The percentage change in the option price given a percentage change in the underlying asset price is $$\frac{C_\mathrm{per}}{C_\mathrm{old}}-1\approx0.089=8.9\%$$. The option elasticity was indeed $$\Delta_\mathrm{old}\frac{S_\mathrm{old}}{C_\mathrm{old}}=8.7$$. As you see, this number gives you the relevant percentage number.
• The same holds for delta: $$\frac{\Delta_\mathrm{per}}{\Delta_\mathrm{old}}-1=0.045=4.5\%$$ which is etimated by $$\Gamma_\mathrm{old}\frac{S_\mathrm{old}}{\Delta_\mathrm{old}}\approx4.6$$. Again, you get the percentage number.
• Leaving the elasticities aside, the percentage change in $$\Delta$$ given an absolute change in the price of the underlying asset is $$\frac{\Delta_\mathrm{per}}{\Delta_\mathrm{old}}-1=0.37$$ which we expected to be $$\frac{\Gamma_\mathrm{old}}{\Delta_\mathrm{old}}\approx 0.46$$.
• The absolute change in $$\Delta$$ given a percentage change in the price of the underlying asset is $$0.027=2.7\%$$ and is estimated to be $$\Gamma_\mathrm{old}S_\mathrm{old}\approx2.74$$. So again, we need to remember the unit.

So, the bottom line is that you have four possibilities (abs|abs (this is gamma), abs|per, per|abs and per|per (elasticity), where per|abs means a percentage change in delta given an absolute change in the price of the underlying asset etc.). Whenever you compute (absolute or relative) changes given a percentage change (i.e. abs|per and per|per), then you need to remember the unit and thus the factor $$\frac{1}{100}$$. In any case, the magnitude of your result ought to indicate whether your result is a percentage or not.

• Why is the practical gamma formula posted as an answer, which has the units of change in delta per % change in asset price, the same as yours but divided by 100? If you write a Taylor's expansion of the delta and wanted to calculate the change in delta for a percent change in asset price dividing by 100 makes sense, but your formula also makes sense so I'm unsure why there's a difference Nov 30, 2019 at 2:19
• @Slade Thank you for your question. I really should've responded earlier. Please see the edit to the answer above. I hope this clarifies the factor $\frac{1}{100}$. Let me know if this answers your question. Dec 1, 2019 at 10:58
• yeah the edit was very thorough. Everything is clear. Thanks a lot! Dec 3, 2019 at 6:24

The traders or practitioners’ gamma concept tries to capture the same issue. It is defined as S times gamma divided by 100:

$$\Gamma_P=\frac{S\, \Gamma}{100}$$